The potential difference across the Coolidge tube is 20 kV and 10 mA current flows through the voltage supply. Only 0.5 % of the energy carried by the electrons striking the target is converted into X-rays. The power carried by X-ray beam (in W) is.

Question

The potential difference across the Coolidge tube is 20 kV and 10 mA current flows through the voltage supply. Only 0.5 % of the energy carried by the electrons striking the target is converted into X-rays. The power carried by X-ray beam (in W) is. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Power of electron beam is P=Vi=20 × 103 × 10 × 10-3 ⇒ P=200 W ⇒ PX- text nay =(0.5/100) × 200=1 W

Q. The potential difference across the Coolidge tube is $20 kV$ and $10 m A$ current flows through the voltage supply. Only $0.5%$ of the energy carried by the electrons striking the target is converted into $X$ -rays. The power carried by $X$ -ray beam (in $W$ ) is___.

Power of electron beam is $P = V_{i} = 20 \times 1 0^{3} \times 10 \times 1 0^{- 3}$
$\Rightarrow P = 200 W$
$\Rightarrow P_{X - nay} = \frac{0.5}{100} \times 200 = 1 W$

Q. The potential difference across the Coolidge tube is 20kV and 10mA current flows through the voltage supply. Only 0.5% of the energy carried by the electrons striking the target is converted into X-rays. The power carried by X-ray beam (in W) is___.

Power of electron beam is P=Vi​=20×103×10×10−3 ⇒P=200W ⇒PX− nay ​=1000.5​×200=1W

Q. The potential difference across the Coolidge tube is $20 kV$ and $10 m A$ current flows through the voltage supply. Only $0.5%$ of the energy carried by the electrons striking the target is converted into $X$ -rays. The power carried by $X$ -ray beam (in $W$ ) is___.

Power of electron beam is $P = V_{i} = 20 \times 1 0^{3} \times 10 \times 1 0^{- 3}$
$\Rightarrow P = 200 W$
$\Rightarrow P_{X - nay} = \frac{0.5}{100} \times 200 = 1 W$